Question

4.
3+/6
√3+√2
= a + b V3
5. Find a and b if​

Answer 1

Answer:

question is not clear please resend

Answer 2

$\bf \frac{3 + \sqrt{6} }{ \sqrt{3} + \sqrt{2} } = a + b \sqrt{3}$

• To figure out the value of a and b we have to rationalize the LHS,

Rationalizing the LHS, we get,

$= \frac{3 + \sqrt{6} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ = \frac{3 + \sqrt{6}( \sqrt{3} - \sqrt{2} ) }{ ( \sqrt{3} + \sqrt{2} ) ( \sqrt{3} - \sqrt{2} ) }$

• In the denominator, an identity is applicable, (a+b)(a-b) = a² - b²

$= \frac{3( \sqrt{3} - \sqrt{2} ) + \sqrt{6} ( \sqrt{3} - \sqrt{2} ) }{ {( \sqrt{3} )}^{2} - { (\sqrt{2}) }^{2} } \\ \\ = \frac{3 \sqrt{3} - 3 \sqrt{2} + \sqrt{18} - \sqrt{12} }{3 - 2} \\ \\ = \frac{3 \sqrt{3} + 3 \sqrt{2} + 3 \sqrt{2} - 2 \sqrt{3} }{1} \\ \\ = 3 \sqrt{3} + 3 \sqrt{2} + 3 \sqrt{2} - 2 \sqrt{3} \\ = (3 - 2)\sqrt{3} + (3 + 3) \sqrt{2} \\ = \sqrt{3} + 6 \sqrt{2} \: \: \: \rm \: or \: \: \red{ 6 \sqrt{2} + \sqrt{3} }$

Thus,

• a = 6√2 & b = 1